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What are the rings whose countable power has the property that every finitely generated submodule is free?

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    $\begingroup$ Are your rings commutative with unity? $\endgroup$ Commented Jun 12 at 23:05
  • $\begingroup$ @Alex Kruckman I would start with a commutative rings with unity. Once that becomes clarified, one can look at the more general case(s) $\endgroup$
    – Rado
    Commented Jun 15 at 3:46

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I think your condition is equivalent to being a semifir (but see the remark below). See Chapter 2 of Paul Cohn, "Free ideal rings and localization in general rings". A commutative semifir is a Bézout domain, so in the commutative case your condition is equivalent to being a Bézout domain. This is more general than a PID.

For a semifir, you also need the ranks of finitely generated free modules to be well defined, so your condition might be a bit more general. Of course, in the commutative case, this is automatic.

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  • $\begingroup$ Great! Interestingly (or strangely), I have never seen this book of Cohn. Worth checking it out. Some people use tensoring to define rank, Another way is to imitate vector spaces. $\endgroup$
    – Rado
    Commented Jun 16 at 22:12
  • $\begingroup$ Thanks. I will check Cohn's book, not sure I have it. $\endgroup$
    – Rado
    Commented Jun 16 at 22:24
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Assuming rings are commutative with unity, these are exactly the Bézout domains (and the zero ring).

Let $R$ be a non-zero ring. The following are equivalent:

  1. $R$ is a Bézout domain (a domain in which every f.g. ideal is principal).
  2. Every f.g. torsion-free module is free.
  3. Every f.g. submodule of $R^\mathbb{N}$ is free.
  4. Every f.g. ideal is a free module.

$1\implies 2$ is the non-trivial implication. See Pete Clark's answer here. It also follows from Lemma 15.22.7 and Lemma 15.124.8 on the Stacks project.

$2\implies 3$ since $R^\mathbb{N}$ is torsion-free (so every submodule is torsion-free).

$3\implies 4$ since every ideal is a submodule of $R$, and hence a submodule of $R^\mathbb{N}$.

For $4\implies 1$, suppose $R$ is not a Bézout domain. Then $R$ has zero-divisors or a f.g. non-principal ideal. If $a\in R$ is a zero-divisor, then $(a)$ is a finitely generated ideal which is not a free module because it has torsion. If $I\subseteq R$ is a f.g. non-principal ideal, then $I$ is not a free module. Indeed, if $a,b\in I$ are non-zero, then $ba-ab=0$, so $I$ has rank at most $1$, but it is not principal, hence not free.

As you can see, there is nothing at all special about $R^{\mathbb{N}}$ being used in the proof. You can replace $R^{\mathbb{N}}$ with any other non-zero torsion-free module, and the equivalence holds.

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  • $\begingroup$ @DaveBenson Yes, thanks. Fixed. $\endgroup$ Commented Jun 16 at 12:05
  • $\begingroup$ Interesting, we can then simply require the condition to be applied to R. $\endgroup$
    – Rado
    Commented Jun 16 at 22:26

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